java - Android:如何每 10 秒通过服务发送一次 http 请求

Question

我需要每 10 秒从服务器接收一次状态。

我尝试通过服务发送 http 请求来做到这一点。

问题是我的代码只执行一次。

这是我的服务代码:

public class ServiceStatusUpdate extends Service {

@Override
public IBinder onBind(Intent intent) {
    // TODO Auto-generated method stub
    return null;
}

@Override
public int onStartCommand(Intent intent, int flags, int startId) {
    while(true)
    {
        try {
            Thread.sleep(10000);
        } catch (InterruptedException e) {
            new DoBackgroundTask().execute(Utilities.QUERYstatus);
            e.printStackTrace();
        }
        return START_STICKY;            
    }
}

private class DoBackgroundTask extends AsyncTask<String, String, String> {

    @Override
    protected String doInBackground(String... params) {
        String response = "";
        String dataToSend = params[0];
        Log.i("FROM STATS SERVICE DoBackgroundTask", dataToSend);
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(Utilities.AGENT_URL);

        try {
            httpPost.setEntity(new StringEntity(dataToSend, "UTF-8"));

            // Set up the header types needed to properly transfer JSON
            httpPost.setHeader("Content-Type", "application/json");
            httpPost.setHeader("Accept-Encoding", "application/json");
            httpPost.setHeader("Accept-Language", "en-US");

            // Execute POST
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity responseEntity = httpResponse.getEntity();
            if (responseEntity != null) {
                response = EntityUtils.toString(responseEntity);
            } else {
                response = "{\"NO DATA:\"NO DATA\"}";
            }
        } catch (ClientProtocolException e) {
            response = "{\"ERROR\":" + e.getMessage().toString() + "}";
        } catch (IOException e) {
            response = "{\"ERROR\":" + e.getMessage().toString() + "}";
        }
        return response;
    }

    @Override
    protected void onPostExecute(String result) {
        Utilities.STATUS = result;
        Log.i("FROM STATUS SERVICE: STATUS IS:", Utilities.STATUS);
        super.onPostExecute(result);
    }
}
}

非常感谢视频

Answer 1

在 onPostExecute 中放置一个处理程序以在 10 秒后发送一个 http 请求

new Handler().postDelayed(new Runnable() {
        public void run() {
            requestHttp();
        }
    }, 10000);

10 秒后 doInBackground 将再次执行，之后再次执行 onPostExecute，再次处理程序，依此类推。